The acceleration due to gravity on the surface of the moon is 1.7 ms^(-2). What is the time period of a simple pendulum on the surface of moon if its time period on the surface earth is 3.5 s. ? (g on the surface earth 9.8 ms^(-2).)

The acceleration due to gravity on the surface of the moon is 1.7 ms^(

1.	The acceleration due to gravity on the surface of the moon is 1.7 ms^(-2). What is the time period of a simple pendulum on the surface of moon if its time period on the surface earth is 3.5 s. ? (g on the surface earth 9.8 ms^(-2).)
Answer» SOLUTION :`g_(m)= 1.7ms^(-2), g_(e)= 9.8 MS^(-1) , T_(e)= 3.5 s^(-1)` As `T_(e)= 2pisqrt((L)/(g_(e)))` and `T_(m)= 2pisqrt((l)/(g_(m)))` `:. (T_(m))/(T_(e))= sqrt((g_(e))/(g_(m)))` or `T_(m)= T_(e) sqrt((g_(e))/(g_(m)))= 3.5 sqrt((9.8)/(1.7))= 8.4 s`